How do I find the length of the cardioid #r=1+sintheta#? Precalculus Polar Coordinates Cardioid Curves 1 Answer Konstantinos Michailidis Nov 7, 2015 The length of the cardiod #r=1+sintheta# between a and b is #L= int_a^b (sqrt(r^2+((dr)/(d(theta)))^2))(d(theta))# hence #dr/(d(theta))=costheta# and #L=int_a^b (sqrt(1+sin^2theta+2costheta+cos^2theta))d(theta)=> L=sqrt2*int_a^b sqrt(1+costheta)d(theta)=> L=sqrt2[(2sqrt(1+costheta))*tan(theta/2)]_a^b=> L=2sqrt2[sqrt(1+cosb)*cos(b/2)-sqrt(1+cosa)*cos(a/2)]# Answer link Related questions What is a cardioid curve? What is the graph of # r = a ± a cos θ#? What is the graph of #r = 2a(1 + cosθ)#? What is the graph of the Cartesian equation #(x^2 + y^2 - 2ax)^2 = 4a^2(x^2 + y^2)#? What is the graph of #r = sin^2(π/8 - θ/4)#? What is the graph of the Cartesian equation #y = 0.75 x^(2/3) +- sqrt(1 - x^2)#? How do I find the area inside a cardioid? How do I find the area inside the cardioid #r=1+costheta#? How do I find the length of the cardioid #r=1+costheta#? How do I graph cardioid #r = 2 + 2cosθ#? See all questions in Cardioid Curves Impact of this question 7529 views around the world You can reuse this answer Creative Commons License